Question: Let $f(x) = 9x^{2}+10x-10$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )?
Answer: The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $9x^{2}+10x-10 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = 9, b = 10, c = -10$ $ x = \dfrac{-10 \pm \sqrt{10^{2} - 4 \cdot 9 \cdot -10}}{2 \cdot 9}$ $ x = \dfrac{-10 \pm \sqrt{460}}{18}$ $ x = \dfrac{-10 \pm 2\sqrt{115}}{18}$ $x =\dfrac{-5 \pm \sqrt{115}}{9}$